# SOLVED: Evaluate the integral by converting it to polar coordinates: âˆ«âˆ« x dy dx Note: Do this problem in steps. 1. Draw a picture of the domain to restate the limits of integration. 2. Change the

Integration by change of variables
Integration by change of variables

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Evaluate the integral by converting it to polar coordinates:
âˆ«âˆ« x dy dx
Note: Do this problem in steps.
1. Draw a picture of the domain to restate the limits of integration.
2. Change the differentials (to match the limits of integration).
3. Use algebra and substitution to change the integrand.
Use polar coordinates to set up and evaluate the double integral âˆ«âˆ« E f(x,y) dA, where f(x,y) = e^(-x^2-y^2) and R: x^2 + y^2 < 25, x > 0.

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05:19

the integral polar coordinates (10 points) Convert to double integral in (x2 + y2)2 dy dxShow all work: Hence_ evaluate the polar double integral:

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Double integrals- – transformation given To evaluate the fol. lowing integrals carry out these steps.a. Sketch the original region of integration $R$ in the $x y$ -plane and the new region $S$ in the uv-plane using the given change of variables.b. Find the limits of integration for the new integral with respect to u and $v.$c. Compute the Jacobian.d. Change variables and evaluate the new integral.$\iint_{R} x y d A,$ where $R$ is bounded by the ellipse $9 x^{2}+4 y^{2}=36$ use $x=2 u, y=3 v$

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Consider the following integral: âˆ«âˆ«(3-y)^2 dxdy + âˆ«âˆ«f dxdySketch the region of integration.Change the order of integration and hence evaluate the integral.

Use polar coordinates to evaluate the following double integral: âˆ«âˆ«r dÎ¸dr / âˆš(r^2 + rcosÎ¸)

Find the Jacobian of the following transformation: r = âˆš(W) and y = âˆš(uV).Sketch the following region:R: 1 < u < 2, 1 < uv < 2 in the UV-plane.Using the above transformation, evaluate the following integral: âˆ«âˆ«âˆš(r^2 + y^2) dydx over the region R.

05:02

Evaluate the double integral in polar coordinates.(x+ydydx by changing it into an equivalent integral

07:59

Double integrals- – transformation given To evaluate the fol. lowing integrals carry out these steps.a. Sketch the original region of integration $R$ in the $x y$ -plane and the new region $S$ in the uv-plane using the given change of variables.b. Find the limits of integration for the new integral with respect to u and $v.$c. Compute the Jacobian.d. Change variables and evaluate the new integral.$\iint_{R} x^{2} y d A,$ where $R=\{(x, y): 0 \leq x \leq 2, x \leq y \leq x+4\}$use $x=2 u, y=4 v+2 u$

Transcript

Hi, here for the given question for the first part we need to evaluate the integral over 0 to a integral over 0 to under root of a square minus x square x dy dx. So here in our case now further we know that the value of integrating this term with respect to y and simplifying we have integration over 0 to a y multiplied with x to the power 0 to under root of a square minus x square dx. So here now on simplifying further we have integration over 0 to a this is equal to under root of x square a square minus x square multiplied with x and here we have dx. So this can be further written as integration over 0 to a 1 by 2 times of under root of a square minus x square multiplied with 2x dx here if we take minus then this will be minus 2. So here now as we can see that this is the derivative of this and simplifying further we have minus 1 by 2 times of under root of a square minus x square. Here we have whole value as cube and here the limit is from 0 to a. So substituting the value of the limit and simplifying this we can say that we got our required value as a square to the power 3 by 2 divided by 3. So this is equal to a cube upon 3. So this is our required solution of the first part. Now here in the second part we are given that here we have value of the function f of x, y is equal to e to the power minus x square plus y square upon 2 and here the region R equals to x square plus y square less than or equal to 25 and x greater than 0. So here now changing into polar coordinate, we have x square…

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